a) is correct
I'd be interested in seeing your work for (b). I don't know how you arrived at that answer.
Consider a infinitesimal cross section of the solid between $(x,x+dx)$
It's a circular disk of radius $\sqrt{x}$ and thickness $dx$
Thus it's differential volume is
$dV = \pi (\sqrt{x})^2~dx = \pi x~dx$
$V = \displaystyle \int_0^9~dV = \displaystyle \int_0^9~\pi x = \left . \pi \dfrac{x^2}{2} \right|_0^9 = \dfrac{81\pi}{2}$
Ok.
Consider a cross section of the volume at height $z$ with thickness $dz$
This circular disk has radius $r=z^2$ and $z \in [0,3]$
$dV = \pi r^2~dz = \pi z^4~dz$
$V = \displaystyle \int_0^3 dV = \int_0^3 \pi z^4 ~dz = \dfrac{243\pi}{5}$
So you were right all along. My apologies.